964: For Polynomials Ring over Field, Irreducibles Are Nonconstants That Can Be Factorized Only with Only 1 Nonconstant

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description/proof of that for polynomials ring over field, irreducibles are nonconstants that can be factorized only with only 1 nonconstant

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of polynomials ring over commutative ring.
• The reader knows a definition of field.
• The reader knows a definition of irreducible element of commutative ring.
• The reader admits the proposition that for the polynomials ring over any field, the units are the nonzero constants.

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Target Context

• The reader will have a description and a proof of the proposition that for the polynomials ring over any field, the irreducibles are the nonconstants that can be factorized only with only 1 nonconstant.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$F[x]$: $=\text{ the polynomials ring over }F$
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Statements:
$\{\text{ the irreducibles on }F[x]\}=\{p(x)\in \{\text{ the nonconstants on }F[x]\}|p(x)=p^{\prime }(x)p^{″}(x)⟹(p^{″}(x)\in \{\text{ the constants on }F[x]\})\vee (p^{\prime }(x)\in \{\text{ the constants on }F[x]\})\}$
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2: Note

If you are wondering how that Statements amounts to "the nonconstants that can be factorized only with only 1 nonconstant", supposing $p(x)=p^{\prime }(x)p^{″}(x)p^{‴}(x)$, if there were 2 nonconstant factors, say $p^{\prime }(x)$ and $p^{‴}(x)$, $p(x)=(p^{\prime }(x)p^{″}(x))p^{‴}(x)$, and $p^{\prime }(x)p^{″}(x)$ or $p^{‴}(x)$ would not be any constant, a contradiction against the Statements; if there was no constant factor, $p(x)=(p^{\prime }(x)p^{″}(x))p^{‴}(x)$, and $p^{\prime }(x)p^{″}(x)$ or $p^{‴}(x)$ would not be any constant, a contradiction against the Statements.

Compare with the proposition that for the polynomials ring over any integral domain, any irreducible can be factorized with at most 1 nonconstant.

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3: Proof

Whole Strategy: Step 1: let $p(x)\in F[x]$ be any irreducible, suppose that $p^{\prime }(x)$ and $p^{″}(x)$ were some nonconstants, find a contradiction; Step 2: let $p(x)\in \{p(x)\in \{\text{ the nonconstants on }F[x]\}|p(x)=p^{\prime }(x)p^{″}(x)⟹(p^{″}(x)\in \{\text{ the constants on }F[x]\})\vee (p^{\prime }(x)\in \{\text{ the constants on }F[x]\})\}$, and see that $p(x)$ is irreducible.

Step 1:

Let $p(x)\in F[x]$ be any irreducible.

$p(x)$ is a nonconstant, because any constant is a unit, by the proposition that for the polynomials ring over any field, the units are the nonzero constants, and any unit is not any irreducible, by the definition of irreducible element of commutative ring.

Let us suppose that $p(x)=p^{\prime }(x)p^{″}(x)$.

Let us suppose that $p^{\prime }(x)$ and $p^{″}(x)$ were some nonconstants. $p^{\prime }(x)$ or $p^{″}(x)$ would not be any unit, by the proposition that for the polynomials ring over any field, the units are the nonzero constants. Then, $p(x)$ would not be any irreducible, by the definition of irreducible element of commutative ring, a contradiction.

So, $p^{\prime }(x)$ or $p^{″}(x)$ is a constant.

That means that $p(x)\in \{p(x)\in \{\text{ the nonconstants on }F[x]\}|p(x)=p^{\prime }(x)p^{″}(x)⟹(p^{″}(x)\in \{\text{ the constants on }F[x]\})\vee (p^{\prime }(x)\in \{\text{ the constants on }F[x]\})\}$.

By the way, when $p^{″}(x)$ is a constant, $p^{\prime }(x)$ is a nonconstant, because otherwise, $p(x)=p^{\prime }(x)p^{″}(x)$ would not be nonconstant, a contradiction; when $p^{\prime }(x)$ is a constant, $p^{″}(x)$ is a nonconstant, because otherwise, $p(x)=p^{\prime }(x)p^{″}(x)$ would not be nonconstant, a contradiction.

Step 2:

Let $p(x)\in \{p(x)\in \{\text{ the nonconstants on }F[x]\}|p(x)=p^{\prime }(x)p^{″}(x)⟹(p^{″}(x)\in \{\text{ the constants on }F[x]\})\vee (p^{\prime }(x)\in \{\text{ the constants on }F[x]\})\}$ be any.

$p(x)\ne 0$.

As $p(x)$ is a nonconstant, $p(x)$ is not any unit, by the proposition that for the polynomials ring over any field, the units are the nonzero constants.

Whenever $p(x)=p^{\prime }(x)p^{″}(x)$, $p^{″}(x)$ is a constant or $p^{\prime }(x)$ is a constant. But for the former case, $p^{″}(x)$ is a unit, by the proposition that for the polynomials ring over any field, the units are the nonzero constants, and for the latter case, $p^{\prime }(x)$ is a unit, by the proposition that for the polynomials ring over any field, the units are the nonzero constants.

So, $p(x)$ is an irreducible, by the definition of irreducible element of commutative ring.

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References

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